Course: SCH 4U
Date: May 20
Topic Being Covered: Calculations involving acidic solutions
Pages in text covered: pg 512-521
Review: Acids, Ka, Kb
Material:

Strong acids dissociate fully. So a 1Mol /L HCl is actually 1
Mol/L of both H+ and Cl-


Must check if the auto-ionization of water has any effect.
For strong acids it does not because usually they are closer to
1 or 2 and the auto-ionization of water only add 1x 10-7 [H+]
ions, which is insignificant
Eg/ what is the pH and pOH of a solution of HNO3 that is 0.25 Mol in
1000mL of solution.
 c = n/v = 0.25 Molar.
 strong acid dissociates fully so [H+] = 0.25
 -log(.25) = 0.6 = pH
 pOH = 13.4
 making at a very strong acid and a very weak base
If HCl is 0.07M, what is the [OH-] ANS = 1.43 x 10-13
 -log (.07) = 1.1549 = pH
 14- 1.1549 = 12.8450 = pOH
 10-pOH = 1.43 x 10-13
What is the pH and pOH of a solution of HBr that is 0.07M in 2L?
ANS: pH = 1.46, pOH = 12.54
Weak acids are a completely different story because they do not fully
ionize!
 If you know the Ka value you can calculate pH
 Before calculations we also need to know the percentage
ionization which tells you how much of the acid will become
products.
 for example if the percentage ionization is 15% then if we
have HA  H+ + A85% will stay HA, and 15% will become H+ + A-
Key points:
Percentage ionization =
[H+]
[HA]
x 100%
So [H+] = (Percent ionization /100) X [HA]
If you know the pH of a weak acid solution, you can figure out the
percentage ionization of the acid.
 for these calculations we will need ice charts
 this allows you to find percentage ionization
 also allows you to find Ka
try this.
If a solution of a weak acid has a pH of 1.92 and we start with 0.2M,
what is the percentage ionization, and what is the Ka value?
 pH of 1.92 = [H+] of 10-1.92 = .012
 Percentage ionization =
[H+]
x 100%
[HA]
 [.012]/[0.2] x 100 = 6% ionization
Ka = Need ice chart
HA
0.2
-.012
.188
I
C
E
Ka = (0.012)2
(.188)
H+
0
+.012
.012
A0
+.012
.012
= 7.65 x 10-4
Eg2/ what is the Ka of a weak acid that has a percentage ionization of
1.3 and starts at 0.1Molar.
HA
0.1
-.0013
.0987
I
C
E

ANS: Ka = 1.7 x 10-5
H+
0
+.0013
.0013
A0
+.0013
.0013
To calculate the pH of weak acid solution
 First we must know Ka and 1 initial concentration.
 set up Ka equation
 fill in ice chart with known information to find variables for ka
equation
 solve for Ka equation
eg/ What is the pH of a solution that has a Ka of 6.6 x10-4 when we
start with a 1Mol/L solution of it.
New idea for solving equations = 100 rule!
 if the concentration of your acid (HA) is over 100 times the
value of the Ka, you can disregard “X” values that would get
added to or subtracted from the HA value at equilibrium
Set up ice chart.
HA
1
-x
1-x
I
C
E
Ka =
H+
0
x
x
A0
x
x
x2
1-x
6.6 x 10-4
x2
1-x
=
= 1/ 6.6 x 10-4 = 1515 = yes it counts. skip X
[HA] / Ka
6.6 x 10-4
6.6 x 10-4
√6.6 x 10-4
does 100 rule count?
x2
1
=
 if not 1, do not forget to multiply out!
x2
=
= x
X = .026 = [H+]
-log(.026) = 1.59  this is the pH of the solution.
Calculating Ka from pH
 take the pH value and make it a H+ equilibrium concentration
 determine Eq concentrations for other species
 plug into Ka formula
 achieve Ka
eg/ A 0.1Mol solution of a weak acid has a pH of 4.23, what is the Ka
of this solution.
ANS = 3.45 x 10-8
Daily extra:
Sample question for test
Assessment
For:
As:
Of:
Descriptive feedback, coaching for improvement.
Goal setting, progress monitoring, determining next steps, self- reflection on thinking and learning.
Show level of understanding and level of knowledge
Assessment used:
Homework: 514 sample 1, 516 practice 1+2 , 518 sample problem 2. pg 520 practice 1 + 2, pg 521
practice 1+2
Notes: